4

I got this from the thread Why are conditionals with false antecedents considered true? And this is what Stefan Perko answered.

Suppose it is raining and it is not raining. Then it is raining. Hence, it is raining OR the moon is made of cheese (1). Since it is not raining, the moon is made of cheese (2).

If you accept:

  1. disjunction introduction, i.e. P -> (P or Q)
  2. disjunctive syllogism, i.e. ((P or Q) and not P) -> Q then you need to accept that contradictory assumptions imply everything.

Also there is a neat mathematical reason, why contradiction "should" imply everything. You can put a (pre)-order on the set of propositions by saying P is less then or equal to Q if and only, if P => Q. Then contradiction is the minimum in this ordered set, because it is less then every other proposition

The first part seems pretty straightforward except how anything can be true and false at the same time seems like a big no-no in logic.

The second quoted part about why contradiction "should" imply everything I can't really follow. Which is the preorder and on what propositions? Isn't it arbitrary to come up with the statement "P is less then or equal to Q? And why is contradiction minimum in this ordered set?

  • 1
    The quote explicitly says what the preorder is, it is preorder by implication. The more a propositions implies the "lesser" it is in it. That contradiction is a minimum is just a rephrasing of it implying everything. This is "neat" in the sense that it makes a natural preorder on propositions have minimal elements, and there is no more to the "should" than having this "neatness". – Conifold Nov 27 '17 at 0:36
  • The first part of the argument is purely "sintactical": it does not rely on the "meaning" (i.e. truth value) of the contradiction, but only on the formal rules of deduction: 1) first premise P; 2) derive P or Q by disjunction intro; 3) second premise not-P; 4) Q from 2) and 3) by disjunctive syllogism. Conclusion: if we agree on some intuitively correct rules of inference, we are forced to conclude that from contradictory premises anything follows (i.e. can be deduced). – Mauro ALLEGRANZA Nov 27 '17 at 10:01
  • @Conifold Could you clarify and give and example. I'm new to this and don't have real grasp of the terms used in explaining things so its a bit confusing. And this concept of a contradiction implies any thing how does it relate to an implication with false antecendent being considered true? Are these two different thing? – MattaCuna Nov 27 '17 at 23:53
  • "Contradiction" is a term of formal logic, where the focus is on formal validity not truth and falsehood, they come up in interpretations. "Sky is blue" may or may not be true, depending on which planet we interpret it, but "sky is blue and sky is not blue" is false in any interpretation because it is a contradiction. From this perspective, yes, since contradiction is always false and implication with false antecedent is always true contradiction has to imply anything. But usually people try to justify it in terms of other, more intuitive, formal rules rather than externally like this. – Conifold Nov 28 '17 at 0:09
  • 1
    A conditional P → Q with false antecedent (i.e. P) is true. A contradiciton P ∧ ¬P is always false; thus, a conditional (P ∧ ¬P) → Q with contradictory antecedent is always true. – Mauro ALLEGRANZA Nov 28 '17 at 8:16
4

I think the proposal goes like so.

Mathematically, a proposition may be seen as a set of possible worlds. Intuitively, this is the set of possible worlds in which the proposition is true. (This allows neat definitions of boolean connectives. For instance, the conjunction of two propositions p and q is just their intersection.) On this picture, a contradiction is the empty set, because it is not true in any possible world. A tautology then is the set of all possible worlds.

Given this, consider the preorder of set inclusion. What does it mean that p is included in q? That all possible worlds in p are also in q. Intuitively, this means that in any possible world where p is true, also q is true – which seems like an adequate understanding of implication. So the story goes: Just as intersection is conjunction, set inclusion is implication.

However, mathematically, the empty set is a subset of any set. By the analogy, this means that a contradiction implies any proposition.

  • Isn't the inclusion relation here the wrong way around for doing the work you want it to do? If A then B gives that B is included in A. One the one hand you have the contradiction implying everything and on the other hand being included by everything. Perhaps I'm missing something. – Steve Lovell Nov 30 '17 at 18:33
  • @SteveLovell There are two notions of "inclusion" here that one might easily confuse. It is true that intuitively, one often says that if B follows from A, then B is somehow already "included" in A, and relatedly that deductions just extract information that is implicitly present in the premises. Set-theoretically, however, this means that the set of worlds in which A is true is included in the set of worlds in which B is true, which indeed is the other way around. – user26652 Nov 30 '17 at 19:54
  • Thanks @DeeDuu, I think my mental picture kept flipping between the two kinds of "inclusion" without my noticing. Upvoting your answer now. – Steve Lovell Nov 30 '17 at 20:03
1

I think the claim is that every statement is derivable from a contradiction. Take the contradiction p & -not-p and any arbitrary statement, q. Informally :

  1. p & not-p (self-contradiction)

  2. p (from 1.)

  3. not-p (from 1.)

  4. not-p v q [not-p or q where 'q' is any arbitrary statement] (or-introduction)

_______________________________________________________________________________]

  1. Therefore q (disjunctive elimination from 2. and 4.

Source : adapted from Lewis & Langford, 'Symbolic Logic', 250.

  • 1
    It seems that you have simply rephrased Perko's answer quoted above by the OP... – Mauro ALLEGRANZA Nov 27 '17 at 12:59
  • &Mauro ALLEGRANZA. Do you mean intentionally rephrased using Perko as a source ? I did no such thing. – Geoffrey Thomas Nov 29 '17 at 13:58
0

The proof found on the example uses a formal artifice, not difficult to follow. Anyway, to simplify it, you can express it as this "if it rains, the moon is made of cheese". Since on a contradiction you have that false=true, you can use the true side: "true, then, the moon is made of cheese". Of course, you can justify any proposition based on such logic.

Isolated, this fallacy seems a trivial game. Nevertheless, if it appears from a result of some kind of normative scheme, it becomes really dangerous. For example, it raises on most religions, since religious books are full of contradictions. Then, you can justify murder, rape or whatever behavior with a simple "God is love for the others, and who does not believe in him will suffer punishment" (fallacy: love is contradictory with punishment).

See ex contradictione quodlibet on Wikipedia.

0

There are two parts to this question.


First part:

The first part seems pretty straightforward except how anything can be true and false at the same time seems like a big no-no in logic.

I agree with the answer provided by Geoffrey Thomas. This is only an alternate way to explain it.

Consider the following proof. Think of "P" in the proof as "It is raining." Think of not "P", or "¬P", in the proof as "It is not raining."

enter image description here

In order to show that one can get anything from a contradiction, we assume we have a contradiction. That is what is happening on line 1. Both "It is raining" and "It is not raining" or "P ∧ ¬P" we assume to be true.

What follows from assuming we have a contradiction on line 1? The claim is that anything should follow. How do we represent anything? Just give it a name and don't specify it. Here I am saying "Q" will stand for "anything". I want to derive anything, or "Q", from the contradiction in line 1, "P ∧ ¬P".

The steps show how this is done.

On lines 2 and 3, I split out the two sentences connected by "and" (∧). On line 2, I put "P". On line 3, I put "¬P". The two sentences on either side of an "and" sentence are called conjuncts and the full sentence is called a conjunction. When I take the two conjuncts and put them on their own lines, I eliminate the conjunction. This is written with the rule "∧E" and they both refer to line 1.

On line 4, I take the "P" on line 2 and realize that if "P" is true it doesn't matter what I "or" with "P". It could be anything. I have a symbol for anything. I called it "Q", so I write "P" or "Q" on line 4 as "P ∨ Q". This or-ing process is called a disjunction and I have just introduced a disjunction on line 4, so I write the rule "∨I" and reference line 2 since that is the line "P" is on.

On line 5, I get the result I am looking for, because I don't have only "P", I also have "¬P" from line 3 which came from the contradiction I assumed on line 1. That means "P" is false (if I look at it from the perspective of "¬P"). Forget that on line 2 I thought "P" was true. This is a contradiction we are working with. So if "P" is false and "P ∨ Q" is true, that means "Q" is true. But "Q" was anything. The rule for doing this is called disjunctive syllogism (DS).

That means if I have a contradiction, I can derive anything.


Second part:

The second quoted part about why contradiction "should" imply everything I can't really follow. Which is the preorder and on what propositions? Isn't it arbitrary to come up with the statement "P is less then or equal to Q? And why is contradiction minimum in this ordered set?

I agree with the answer provided by user26652 for this part, but I will try to describe it differently.

Note that Stefan Perko is not trying to argue that having a contradiction implies anything. He's already done that in the first part and he assumes you have accepted it.

Rather he is trying to show that there is an intuitive way of thinking about a contradiction from the perspective of set theory by thinking of a contradiction as something insignificant, even empty, like the empty set. The empty set is a subset of any set (that is, it is the subset of "anything") just like a contradiction implies anything. He is pointing out a similarity between a contradiction and the empty set.

It seems odd to think of a contradiction as something small. There is even a rule called explosion. If we get a contradiction we can use the rule of explosion and derive anything we want as the first part showed.

However, to see how smallness might be a good metaphor for a contradiction rather than a large explosion consider what an implication might look like using Venn diagrams. In the following diagram "P" implies "Q". Whenever we have "P", we have "Q". "Q" is in a sense bigger than "P". Note that "P" represents a possibly smaller circle than "Q" in the diagram. It cannot be a circle bigger than "Q".

enter image description here

Now think of "P" as a contradiction getting smaller and smaller in that diagram and implying anything. Just like the empty set is so small that it is a subset of any set.


Reference

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.